The effusion rates, R, of two gases, A and B, are related to their molecular massesaccording to the formula below. Calculate RB given that RA = 50.0 ml/sec and the molecular masses are MA =40.0andMB = 28.0 amu. Place your answer in the box with unitets and three significant figures.
This will be my second time submitting this question. I tried following the steps that were sent to me, but it seems as if some steps were missing assuming I know the obvious. Anyway, I am aware this happens with some people of today. I need to know every step so I can understand. I may stress this is new and I have to get in the habit of solving it. Also, I should receive credit for this question due to me asking the same question twice.
According to the Graham's law of effusion the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.This formula can be written as:
where:
Rate1 is the rate of effusion of the first gas (volume or number of moles per unit time).
Rate2 is the rate of effusion for the second gas.
M1 is the molar mass of gas 1
M2 is the molar mass of gas 2.
RA/RB =
50/RB =
50/RB =0.836
RB = 50/0.836
RB = 59.8 mol/sec
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